Complex analysis f'/f , f meromorphic, laurent series

1. The problem statement, all variables and given/known data

consider

f

a meromorphic function with a finite pole at

z=a

of order

m

.

Thus

f(z)

has a laurent expansion:

f(z)=\sum\limits_{n=-m}^{\infty} a_{n} (z-a)^{n}

I want to show that

f'(z)'/f(z)= \frac{m}/{z-a} + holomorphic function

And so where a holomporphic function is one where the laurent expansion above will instead start at n=0

2. Relevant equations

So I've tried writing out a few terms in the both f'(z) and f(z) to try and see this, but I don't seem to be going anywhere..

.

f(z)=\sum\limits_{n=-m}^{\infty} a_{n} (z-a)^{n}

, for a meromorphic function with pole at a of order m

[br][br][br][br] f(z)=\sum\limits_{n=0}^{\infty} a_{n} (z-a)^{n}

, for a holomorphic function

3. The attempt at a solution

f(z) = a_{-m} / (t-a)^{m} + a_{-m+1}/(t-a)^{m-1} + a_{-m+2}/(t-a)^{m-2} +...+ a_{-1}/(t-a)+a_{0} + a_{1}(t-a) +...

f'(z) = -m a_{-m} (t-a)^{-m-1} +(-m+1)a_{m+1}(t-a)^{-m} + (-m+2) a_{-m+2} (t-a)^{-m+1} + ... + (-a_{-1} (t-a)^{-2} + 0 + a_{1} + 2a_{2}(t-a)+...

I have no idea how to divide properly.

So I look at the terms with the

a_{-m}

coefficient and see that

-m a_{-m} (t-a)^{-m-1} / a_{-m} / (t-a)^{m} = -m / (t-a)

which looks sort of on track,however....i get the same when i do this with each of the next coefficients of a , i.e. the

(t-a)^{-1}

help greatly appreciated, many thanks.

Reply 1

atsruser

Original post by xfootiecrazeesarax

1. The problem statement, all variables and given/known data

consider

f

a meromorphic function with a finite pole at

z=a

of order

m

.

Thus

f(z)

has a laurent expansion:

f(z)=\sum\limits_{n=-m}^{\infty} a_{n} (z-a)^{n}

I want to show that

f'(z)'/f(z)= \frac{m}/{z-a} + holomorphic function

And so where a holomporphic function is one where the laurent expansion above will instead start at n=0

1. You need to improve your presentation a bit. This was hard to follow and incorrect in at least one place. And I think your quoted result is missing a - sign on the first term.

2. Note that

f(z) = (z-a)^{-m} g(z)

where

g(z)

is analytic. Now form

f'(z)

and bear in mind that the quotient of two analytic functions is analytic.

Reply 2

xfootiecrazeesarax

Original post by atsruser

1. You need to improve your presentation a bit. This was hard to follow and incorrect in at least one place. And I think your quoted result is missing a - sign on the first term.

2. Note that

f(z) = (z-a)^{-m} g(z)

where

g(z)

is analytic. Now form

f'(z)

and bear in mind that the quotient of two analytic functions is analytic.

ahh thanks man !

Reply 3

xfootiecrazeesarax

Original post by atsruser

1. You need to improve your presentation a bit. This was hard to follow and incorrect in at least one place. And I think your quoted result is missing a - sign on the first term.

2. Note that

f(z) = (z-a)^{-m} g(z)

where

g(z)

is analytic. Now form

f'(z)

and bear in mind that the quotient of two analytic functions is analytic.

ahh yes I believe my final result is missing a minus sign,

Reply 4

xfootiecrazeesarax

Original post by atsruser

1. You need to improve your presentation a bit. This was hard to follow and incorrect in at least one place. And I think your quoted result is missing a - sign on the first term.

2. Note that

f(z) = (z-a)^{-m} g(z)

where

g(z)

is analytic. Now form

f'(z)

and bear in mind that the quotient of two analytic functions is analytic.

sorry to re-bump an old thread, but it seemed best suited here.

how is

g'(z)

holomorphic?
I have

g'(z) = f'(z)(z-a)^{m}+m(z-a)^{m-1}f(z)

,

for the pole of

f

to be 'cancelled out' we needed to multiply by

(z-a)^{m}

. The first term however will now have a term

m+1

derivaitve of

z-a

and the second is now multiplied only by

(z-a)^{m-1}

?

thanks

(edited 7 years ago)

Reply 5

atsruser

Original post by xfootiecrazeesarax

sorry to re-bump an old thread, but it seemed best suited here.

how is ##g'(z)## holomorphic

Well, unless I'm confused about what you're asking,

g'(z)

is the derivative of

g(z)

which is holomorphic, and since holomorphic functions are infinitely differentiable, then it's holomorphic.

I have ##g'(z) = f'(z)(z-a)^{m}+m(z-a)^{m-1}f(z)##,

for the pole of ##f## to be 'cancelled out' we needed to multiply by ##(z-a)^{m}##. The first term however will now have a term ##m+1## derivaitve of ##z-a## and the second is now multiplied only by ##(z-a)^{m-1}## ?

thanks

I'm finding it hard to follow this. Could you latex it up a bit and write it out in full mathematically, rather than merely describing it in words?

Reply 6

DFranklin

Original post by xfootiecrazeesarax

sorry to re-bump an old thread, but it seemed best suited here.

how is ##g'(z)## holomorphic?
I have ##g'(z) = f'(z)(z-a)^{m}+m(z-a)^{m-1}f(z)##,g(z) is holomorphic, therefore so are all its derivatives.

for the pole of ##f## to be 'cancelled out' we needed to multiply by ##(z-a)^{m}##. The first term however will now have a term ##m+1## derivaitve of ##z-a## and the second is now multiplied only by ##(z-a)^{m-1}## ?

If you look at the actual series, the term you are worried about is constant. I.e. it's of the form C(z-a)^0. When k = 0, we don't need to worry that the

Unparseable latex formula:

\latex \dfrac{d}{dz} (z-a)^k = k(z-a)^{k-1}

rule for derivatives seems to imply we get a k/(z-a) term - the term is constant and so its derivaitve is 0.

Reply 7

xfootiecrazeesarax

Original post by atsruser

Well, unless I'm confused about what you're asking,

g'(z)

is the derivative of

g(z)

which is holomorphic, and since holomorphic functions are infinitely differentiable, then it's holomorphic.

I'm finding it hard to follow this. Could you latex it up a bit and write it out in full mathematically, rather than merely describing it in words?